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## Section1.2Basic properties of the definite integral

When we studied limits and derivatives, we developed methods for taking limits or derivatives of “complicated functions” like $f(x)=x^2 + \sin(x)$ by understanding how limits and derivatives interact with basic arithmetic operations like addition and subtraction. This allowed us to reduce the problem into one of of computing derivatives of simpler functions like $x^2$ and $\sin(x)\text{.}$ Along the way we established simple rules such as

\begin{align*} \lim_{x\to a}(f(x)+g(x)) &= \lim_{x\to a}f(x) + \lim_{x\to a} g(x) &\text{and}&& \diff{}{x}(f(x)+g(x)) &= \diff{f}{x} + \diff{g}{x} \end{align*}

Some of these rules have very natural analogues for integrals and we discuss them below. Unfortunately the analogous rules for integrals of products of functions or integrals of compositions of functions are more complicated than those for limits or derivatives. We discuss those rules at length in subsequent sections. For now let us consider some of the simpler rules of the arithmetic of integrals.

It is not too hard to prove this result from the definition of the definite integral. Additionally we only really need to prove (d) and (e) since

• (a) follows from (d) by setting $A=B=1\text{,}$
• (b) follows from (d) by setting $A=1, B=-1\text{,}$ and
• (c) follows from (d) by setting $A=C, B=0\text{.}$

As noted above, it suffices for us to prove (d) and (e). Since (e) is easier, we will start with that. It is also a good warm-up for (d).

• The definite integral in (e), $\int_a^b 1 \dee{x}\text{,}$ can be interpreted geometrically as the area of the rectangle with height 1 running from $x=a$ to $x=b\text{;}$ this area is clearly $b-a\text{.}$ We can also prove this formula from the definition of the integral (Definition 1.1.9): \begin{align*} \int_a^b\dee{x} &=\lim_{n\rightarrow\infty}\sum_{i=1}^n f(x_{i,n}^*)\,\frac{b-a}{n} &\text{by definition}\\ &=\lim_{n\rightarrow\infty}\sum_{i=1}^n 1\,\frac{b-a}{n} &\text{since $f(x)=1$}\\ &=\lim_{n\rightarrow\infty}(b-a) \sum_{i=1}^n \frac{1}{n} &\text{since $a,b$ are constants}\\ &=\lim_{n\rightarrow\infty}(b-a)\\ &=b-a \end{align*} as required.
• To prove (d) let us start by defining $h(x) = Af(x)+Bg(x)$ and then we need to express the integral of $h(x)$ in terms of those of $f(x)$ and $g(x)\text{.}$ We use Definition 1.1.9 and some algebraic manipulations  1 Now is a good time to look back at Theorem 1.1.5. to arrive at the result. \begin{align*} \int_a^bh(x) \dee{x} &= \sum_{i=1}^n h(x_{i,n}^*)\cdot\frac{b-a}{n} &\text{by Definition }\knowl{./knowl/def_INTintegral.html}{\text{1.1.9}}\\ &= \sum_{i=1}^n \left(Af(x_{i,n}^*)+Bg(x_{i,n}^*) \right)\cdot \frac{b-a}{n}\\ &= \sum_{i=1}^n \left(Af(x_{i,n}^*)\cdot \frac{b-a}{n} + Bg(x_{i,n}^*)\cdot \frac{b-a}{n} \right)\\ &= \left(\sum_{i=1}^n Af(x_{i,n}^*)\cdot \frac{b-a}{n}\right) + \left(\sum_{i=1}^n Bg(x_{i,n}^*)\cdot \frac{b-a}{n}\right) &\text{by Theorem }\knowl{./knowl/thm_INTsummationArith.html}{\text{1.1.5}}\text{(b)}\\ &= A\left(\sum_{i=1}^n f(x_{i,n}^*)\cdot \frac{b-a}{n}\right) + B\left(\sum_{i=1}^n g(x_{i,n}^*)\cdot \frac{b-a}{n}\right) &\text{by Theorem }\knowl{./knowl/thm_INTsummationArith.html}{\text{1.1.5}}\text{(a)}\\ &= A \int_a^b f(x) \dee{x} + B \int_a^b g(x) \dee{x} &\text{by Definition }\knowl{./knowl/def_INTintegral.html}{\text{1.1.9}} \end{align*} as required.

Using this Theorem we can integrate sums, differences and constant multiples of functions we know how to integrate. For example:

In Example 1.1.1 we saw that $\int_0^1 e^x\dee{x}=e-1\text{.}$ So

\begin{align*} \int_0^1\big(e^x+7\big)\dee{x} &= \int_0^1 e^x\dee{x} + 7\int_0^1 1 \dee{x} \\ & \text{by Theorem }\knowl{./knowl/thm_Intarith.html}{\text{1.2.1}}\text{(d) with $A=1,f(x)=e^x,B=7,g(x)=1$} \\ &=(e-1)+7\times (1-0) \\ &\text{by Example }\knowl{./knowl/eg_INTexparea.html}{\text{1.1.1}}\text{ and Theorem }\knowl{./knowl/thm_Intarith.html}{\text{1.2.1}}\text{(e)} \\ &=e+6 \end{align*}

When we gave the formal definition of $\int_a^b f(x) \dee{x}$ in Definition 1.1.9 we explained that the integral could be interpreted as the signed area between the curve $y=f(x)$ and the $x$-axis on the interval $[a,b]\text{.}$ In order for this interpretation to make sense we required that $a \lt b\text{,}$ and though we remarked that the integral makes sense when $a \gt b$ we did not explain any further. Thankfully there is an easy way to express the integral $\int_a^b f(x)\dee{x}$ in terms of $\int_b^a f(x)\dee{x}$ — making it always possible to write an integral so the lower limit of integration is less than the upper limit of integration. Theorem 1.2.3, below, tell us that, for example, $\int_7^3 e^x\dee{x} = - \int_3^7 e^x\dee{x}\text{.}$ The same theorem also provides us with two other simple manipulations of the limits of integration.

The proof of this statement is not too difficult.

Let us prove the statements in order.

• Consider the definition of the definite integral \begin{align*} \int_a^b f(x) \dee{x} &= \lim_{n \to \infty} \sum_{i=1}^n f(x_{i,n}^*)\cdot\frac{b-a}{n} \end{align*} If we now substitute $b=a$ in this expression we have \begin{align*} \int_a^a f(x) \dee{x} &= \lim_{n \to \infty} \sum_{i=1}^n f(x_{i,n}^*)\cdot\underbrace{\frac{a-a}{n}}_{=0}\\ &= \lim_{n \to \infty} \sum_{i=1}^n \underbrace{f(x_{i,n}^*)\cdot 0}_{=0}\\ &= \lim_{n \to \infty} 0 \\ &= 0 \end{align*} as required.
• Consider now the definite integral $\int_a^b f(x) \dee{x}\text{.}$ We will sneak up on the proof by first examining Riemann sum approximations to both this and $\int_b^a f(x)\dee{x}\text{.}$ The midpoint Riemann sum approximation to $\int_a^b f(x)\dee{x}$ with $4$ subintervals (so that each subinterval has width $\frac{b-a}{4}$) is

\begin{align*} &\left\{f\Big(a+\frac{1}{2}\frac{b-a}{4}\Big) +f\Big(a+\frac{3}{2}\frac{b-a}{4}\Big) +f\Big(a+\frac{5}{2}\frac{b-a}{4}\Big) + f\Big(a+\frac{7}{2}\frac{b-a}{4}\Big) \right\}\cdot\frac{b-a}{4}\\ &=\left\{f\Big(\frac{7}{8}a+\frac{1}{8}b\Big) +f\Big(\frac{5}{8}a+\frac{3}{8}b\Big) +f\Big(\frac{3}{8}a+\frac{5}{8}b\Big) +f\Big(\frac{1}{8}a+\frac{7}{8}b\Big)\right\}\cdot\frac{b-a}{4} \end{align*}

Now we do the same for $\int_b^a f(x)\dee{x}$ with $4$ subintervals. Note that $b$ is now the lower limit on the integral and $a$ is now the upper limit on the integral. This is likely to cause confusion when we write out the Riemann sum, so we'll temporarily rename $b$ to $A$ and $a$ to $B\text{.}$ The midpoint Riemann sum approximation to $\int_A^B f(x)\dee{x}$ with $4$ subintervals is

\begin{align*} &\left\{f\Big(A+\frac{1}{2}\frac{B-A}{4}\Big) +f\Big(A+\frac{3}{2}\frac{B-A}{4}\Big) +f\Big(A+\frac{5}{2}\frac{B-A}{4}\Big) +f\Big(A+\frac{7}{2}\frac{B-A}{4}\Big)\right\}\cdot \frac{B-A}{4}\\ &=\left\{f\Big(\frac{7}{8}A+\frac{1}{8}B\Big) +f\Big(\frac{5}{8}A+\frac{3}{8}B\Big) +f\Big(\frac{3}{8}A+\frac{5}{8}B\Big) +f\Big(\frac{1}{8}A+\frac{7}{8}B\Big)\right\}\cdot \frac{B-A}{4}\\ \end{align*}

Now recalling that $A=b$ and $B=a\text{,}$ we have that the midpoint Riemann sum approximation to $\int_b^a f(x)\dee{x}$ with $4$ subintervals is

\begin{align*} &\left\{f\Big(\frac{7}{8}b+\frac{1}{8}a\Big) +f\Big(\frac{5}{8}b+\frac{3}{8}a\Big) +f\Big(\frac{3}{8}b+\frac{5}{8}a\Big) +f\Big(\frac{1}{8}b+\frac{7}{8}a\Big)\right\}\cdot \frac{a-b}{4} \end{align*}

Thus we see that the Riemann sums for the two integrals are nearly identical — the only difference being the factor of $\frac{b-a}{4}$ versas $\frac{a-b}{4}\text{.}$ Hence the two Riemann sums are negatives of each other.

The same computation with $n$ subintervals shows that the midpoint Riemann sum approximations to $\int_b^a f(x)\dee{x}$ and $\int_a^b f(x)\dee{x}$ with $n$ subintervals are negatives of each other. Taking the limit $n\rightarrow\infty$ gives $\int_b^a f(x)\dee{x}= -\int_a^b f(x)\dee{x}\text{.}$

• Finally consider (c) — we will not give a formal proof of this, but instead will interpret it geometrically. Indeed one can also interpret (a) geometrically. In both cases these become statements about areas:

are

\begin{gather*} \text{Area}\big\{\ (x,y)\ \big|\ a\le x\le a,\ 0\le y\le f(x)\ \big\}=0 \end{gather*}

and

\begin{align*} \text{Area}\big\{\ (x,y)\ \big|\ a\le x\le b,\ 0\le y\le f(x)\ \big\} &=\text{Area}\big\{\ (x,y)\ \big|\ a\le x\le c,\ 0\le y\le f(x)\ \big\} \\ & +\text{Area}\big\{\ (x,y)\ \big|\ c\le x\le b,\ 0\le y\le f(x)\ \big\} \end{align*}

respectively. Both of these geometric statements are intuitively obvious. See the figures below.

Note that we have assumed that $a\leq c \leq b$ and that $f(x)\geq 0\text{.}$ One can remove these restrictions and also make the proof more formal, but it becomes quite tedious and less intuitive.

###### Remark1.2.4

For notational simplicity, let's assume that $a\le c\le b$ and $f(x)\ge 0$ for all $a\le x\le b\text{.}$ The geometric interpretations of the identities

are

\begin{gather*} \text{Area}\big\{\ (x,y)\ \big|\ a\le x\le a,\ 0\le y\le f(x)\ \big\}=0 \end{gather*}

and

\begin{align*} \text{Area}\big\{\ (x,y)\ \big|\ a\le x\le b,\ 0\le y\le f(x)\ \big\} &=\text{Area}\big\{\ (x,y)\ \big|\ a\le x\le c,\ 0\le y\le f(x)\ \big\} \\ & +\text{Area}\big\{\ (x,y)\ \big|\ c\le x\le b,\ 0\le y\le f(x)\ \big\} \end{align*}

respectively. Both of these geometric statements are intuitively obvious. See the figures below. We won't give a formal proof.

So we concentrate on the formula $\int_b^a f(x)\dee{x}= -\int_a^b f(x)\dee{x}\text{.}$ The midpoint Riemann sum approximation to $\int_a^b f(x)\dee{x}$ with $4$ subintervals (so that each subinterval has width $\frac{b-a}{4}$) is

\begin{align} &\Big\{f\Big(a+\frac{1}{2}\frac{b-a}{4}\Big) +f\Big(a+\frac{3}{2}\frac{b-a}{4}\Big) +f\Big(a+\frac{5}{2}\frac{b-a}{4}\Big) +f\Big(a+\frac{7}{2}\frac{b-a}{4}\Big)\Big\}\ \frac{b-a}{4}\notag\\ &=\Big\{f\Big(\frac{7}{8}a+\frac{1}{8}b\Big) +f\Big(\frac{5}{8}a+\frac{3}{8}b\Big) +f\Big(\frac{3}{8}a+\frac{5}{8}b\Big) +f\Big(\frac{1}{8}a+\frac{7}{8}b\Big)\Big\}\ \frac{b-a}{4}\tag{$\star$} \end{align}

We're now going to write out the midpoint Riemann sum approximation to $\int_b^a f(x)\dee{x}$ with $4$ subintervals. Note that $b$ is now the lower limit on the integral and $a$ is now the upper limit on the integral. This is likely to cause confusion when we write out the Riemann sum, so we'll temporarily rename $b$ to $A$ and $a$ to $B\text{.}$ The midpoint Riemann sum approximation to $\int_A^B f(x)\dee{x}$ with $4$ subintervals is

\begin{align*} &\Big\{f\Big(A+\frac{1}{2}\frac{B-A}{4}\Big) +f\Big(A+\frac{3}{2}\frac{B-A}{4}\Big) +f\Big(A+\frac{5}{2}\frac{B-A}{4}\Big) +f\Big(A+\frac{7}{2}\frac{B-A}{4}\Big)\Big\}\ \frac{B-A}{4}\\ &=\Big\{f\Big(\frac{7}{8}A+\frac{1}{8}B\Big) +f\Big(\frac{5}{8}A+\frac{3}{8}B\Big) +f\Big(\frac{3}{8}A+\frac{5}{8}B\Big) +f\Big(\frac{1}{8}A+\frac{7}{8}B\Big)\Big\}\ \frac{B-A}{4} \end{align*}

Now recalling that $A=b$ and $B=a\text{,}$ we have that the midpoint Riemann sum approximation to $\int_b^a f(x)\dee{x}$ with $4$ subintervals is

\begin{gather} \Big\{f\Big(\frac{7}{8}b+\frac{1}{8}a\Big) +f\Big(\frac{5}{8}b+\frac{3}{8}a\Big) +f\Big(\frac{3}{8}b+\frac{5}{8}a\Big) +f\Big(\frac{1}{8}b+\frac{7}{8}a\Big)\Big\}\ \frac{a-b}{4}\tag{$\star\star$} \end{gather}

The curly brackets in ($\star$) and ($\star\star$) are equal to each other — the terms are just in the reverse order. The factors multiplying the curly brackets in ($\star$) and ($\star\star$), namely $\frac{b-a}{4}$ and $\frac{a-b}{4}\text{,}$ are negatives of each other, so ($\star\star$)$=-$($\star$). The same computation with $n$ subintervals shows that the midpoint Riemann sum approximations to $\int_b^a f(x)\dee{x}$ and $\int_a^b f(x)\dee{x}$ with $n$ subintervals are negatives of each other. Taking the limit $n\rightarrow\infty$ gives $\int_b^a f(x)\dee{x}= -\int_a^b f(x)\dee{x}\text{.}$

Back in Example 1.1.14 we saw that when $b \gt 0$ $\int_0^b x\dee{x} =\frac{b^2}{2}\text{.}$ We'll now verify that $\int_0^b x\dee{x} =\frac{b^2}{2}$ is still true when $b=0$ and also when $b \lt 0\text{.}$

• First consider $b=0\text{.}$ Then the statement $\int_0^b x\dee{x} =\frac{b^2}{2}$ becomes \begin{gather*} \int_0^0 x\dee{x} =0 \end{gather*} This is an immediate consequence of Theorem 1.2.3(a).
• Now consider $b \lt 0\text{.}$ Let us write $B=-b\text{,}$ so that $B \gt 0\text{.}$ In Example 1.1.14 we saw that \begin{gather*} \int_{-B}^0 x\dee{x} =-\frac{B^2}{2}. \end{gather*} So we have \begin{align*} \int_0^b x\dee{x} &=\int^{-B}_0 x\dee{x} =- \int_{-B}^0 x\dee{x} & \text{by Theorem }\knowl{./knowl/thm_Intdomain.html}{\text{1.2.3}}\text{(b)}\\ & =-\left(-\frac{B^2}{2}\right) & \text{by Example }\knowl{./knowl/eg_INTtriangle.html}{\text{1.1.14}}\\ & =\frac{B^2}{2} = \frac{b^2}{2} \end{align*}

We have now shown that

\begin{align*} \int_0^b x\dee{x} &=\frac{b^2}{2} &\text{ for all real numbers $b$} \end{align*}

Applying Theorem 1.2.3 yet again, we have, for all real numbers $a$ and $b\text{,}$

\begin{align*} \int_a^b x\dee{x} &= \int_a^0 x\dee{x} + \int_0^b x\dee{x} & \text{by Theorem }\knowl{./knowl/thm_Intdomain.html}{\text{1.2.3}}(c)\text{ with $c=0$}\\ &= \int_0^b x\dee{x} - \int_0^a x\dee{x} & \text{by Theorem }\knowl{./knowl/thm_Intdomain.html}{\text{1.2.3}}\text{(b)}\\ &=\frac{b^2-a^2}{2} & \text{by Example }\knowl{./knowl/eg_INTPROPxa.html}{\text{1.2.5}}\text{, twice} \end{align*}

We can also understand this result geometrically.

• (left) When $0 \lt a \lt b\text{,}$ the integral represents the area in green which is the difference of two right-angle triangles — the larger with area $b^2/2$ and the smaller with area $a^2/2\text{.}$
• (centre) When $a \lt 0 \lt b\text{,}$ the integral represents the signed area of the two displayed triangles. The one above the axis has area $b^2/2$ while the one below has area $-a^2/2$ (since it is below the axis).
• (right) When $a \lt b \lt 0\text{,}$ the integral represents the signed area in purple of the difference between the two triangles — the larger with area $-a^2/2$ and the smaller with area $-b^2/2\text{.}$

Theorem 1.2.3(c) shows us how we can split an integral over a larger interval into one over two (or more) smaller intervals. This is particularly useful for dealing with piece-wise functions, like $|x|\text{.}$

Using Theorem 1.2.3, we can readily evaluate integrals involving $|x|\text{.}$ First, recall that

\begin{align*} |x|=\begin{cases} x & \text{if $x\ge 0$} \\ -x & \text{if $x \lt 0$} \end{cases} \end{align*}

Now consider (for example) $\int_{-2}^3 |x| \dee{x}\text{.}$ Since the integrand changes at $x=0\text{,}$ it makes sense to split the interval of integration at that point:

\begin{align*} \int_{-2}^3 |x| \dee{x} &= \int_{-2}^0 |x| \dee{x} + \int_0^3 |x| \dee{x} &\text{by Theorem }\knowl{./knowl/thm_Intdomain.html}{\text{1.2.3}}\\ &= \int_{-2}^0 (-x) \dee{x} + \int_0^3 x \dee{x} &\text{by definition of $|x|$}\\ &= -\int_{-2}^0 x\dee{x} + \int_0^3 x \dee{x} &\text{by Theorem }\knowl{./knowl/thm_Intarith.html}{\text{1.2.1}}\text{(c)}\\ &= - (-2^2/2) + (3^2/2) = (4+9)/2\\ &= 13/2 \end{align*}

We can go further still — given a function $f(x)$ we can rewrite the integral of $f(|x|)$ in terms of the integral of $f(x)$ and $f(-x)\text{.}$

\begin{align*} \int_{-1}^1 f\big(|x|\big)\dee{x} & = \int_{-1}^0 f\big(|x|\big)\dee{x}+ \int_0^1 f\big(|x|\big)\dee{x} \\ & = \int_{-1}^0 f(-x)\dee{x}+ \int_0^1 f(x)\dee{x} \end{align*}

Here is a more concrete example.

Let us compute $\int_{-1}^1 \big(1-|x|\big)\dee{x}$ again. In Example 1.1.15 we evaluated this integral by interpretting it as the area of a triangle. This time we are going to use only the properties given in Theorems 1.2.1 and 1.2.3 and the facts that

\begin{align*} \int_a^b \dee{x} &= b-a &\text{and}&& \int_a^b x\dee{x}=\frac{b^2-a^2}{2} \end{align*}

That $\int_a^b\dee{x} = b-a$ is part (e) of Theorem 1.2.1. We saw that $\int_a^b x\dee{x}=\frac{b^2-a^2}{2}$ in Example 1.2.6.

First we are going to get rid of the absolute value signs by splitting the interval over which we integrate. Recalling that $|x|=x$ whenever $x\ge 0$ and $|x|=-x$ whenever $x\le 0\text{,}$ we split the interval by Theorem 1.2.3(c),

\begin{align*} \int_{-1}^1 \big(1-|x|\big)\dee{x} &=\int_{-1}^0 \big(1-|x|\big)\dee{x} + \int_0^1 \big(1-|x|\big)\dee{x} \\ &=\int_{-1}^0 \big(1-(-x)\big)\dee{x} + \int_0^1 \big(1-x\big)\dee{x} \\ &=\int_{-1}^0 \big(1+x\big)\dee{x} + \int_0^1 \big(1-x\big)\dee{x} \end{align*}

Now we apply parts (a) and (b) of Theorem 1.2.1, and then

\begin{align*} \int_{-1}^1 \big[1-|x|\big]\dee{x} &=\int_{-1}^0 1\dee{x} + \int_{-1}^0 x\dee{x} + \int_0^1 1\dee{x} - \int_0^1 x\dee{x} \\ &=[0-(-1)]+\frac{0^2-(-1)^2}{2}+[1-0]-\frac{1^2-0^2}{2} \\ &=1 \end{align*}